Iowa State University Capstones, Theses and Retrospective Theses and Dissertations Dissertations
1966 Meet representations in upper continuous modular lattices James Elton Delany Iowa State University
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Recommended Citation Delany, James Elton, "Meet representations in upper continuous modular lattices " (1966). Retrospective Theses and Dissertations. 2894. https://lib.dr.iastate.edu/rtd/2894
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DE LA NY, James Elton, 1941— MEET REPRESENTATIONS IN UPPER CONTINU OUS MODULAR LATTICES.
Iowa State University of Science and Technology Ph.D., 1966 Mathematics
University Microfilms, Inc.. Ann Arbor, Michigan MEET REPRESENTATIONS IN UPPER CONTINUOUS MODULAR LATTICES
by
James Elton Delany
A Dissertation Submitted to the
Graduate Faculty in Partial Fulfillment of The Requirements for the Degree of
DOCTOR OF PHILOSOPHY
Major Subject: Mathematics
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Iowa State University of Science and Technology Ames, Iowa
1966 il
TABLE OP CONTENTS
Page 1. INTRODUCTION 1 2. TORSION ELEMENTS, TORSION FREE ELEMENTS, AND COVERING CONDITIONS 4
3. ATOMIC LATTICES AND UPPER CONTINUITY 8
4. COMPLETE JOIN HOMOMORPHISMS 16
5. TWO HOMOMORPHISMS 27
6. MEET IRREDUCIBLE ELEMENTS 36
7. IRREDUNDANT REPRESENTATIONS 41
8. APPLICATIONS 51
9. BIBLIOGRAPHY 58
10. ACKNOWLEDGMENTS 59 1
1. INTRODUCTION
The study of representations of elements of the lattice of congruence relations of an algebra as intersections of other elements of this lattice is of interest for two reasons. In the first place, such a study offers an extension of the classical representation theory for the ideals of a noetherian ring. Secondly, there is a natural correspondence between the subdirect product representations of an algebra and the meet representations of the zero element of its lattice of con gruence relations. The lattice of congruence relations of any algebra is compactly generated. (An element x of a lattice is compact if whenever x 4US there is a finite set FcS such that x^ijF, A complete lattice is said to be compactly generated if each element is the union of a set of compact elements.)
In order to study the subdirect product representations of an algebra it often suffices to investigate meet representations in compactly generated lattices. For example, the fact that any algebra has a subdirect product representation in terms of subdirectly irreducible factor algebras may be regarded as a consequence of the fact that each element of a compactly generated lattice has a meet representation in terms of com- I pletely meet irreducible elements. (An element x of a lattice is meet irreducible if xssj^o Sg implies that either x=s^ or xaSg. It is completely meet irreducible if x=Os implies X g S.) A meet representation of an element, say x=S, is 2
irredundant if x true that it must possess an irredundant one. For example, Dilworth and Crawley have characterized the compactly generated modular lattices in which each element has such a representa tion as those which are atomic (2). (A lattice is atomic if whenever x will suffice to consider upper continuous modular lattices. (A complete lattice is upper continuous if for each element X and each chain of elements S the relation x A Us=(Jg (xn s) holds.) Any compactly generated lattice is upper continuous. On one hand we will see that there is an analogy between the case dealing with meet irreducible elements and that in volving completely meet irreducible elements. This may be seen by comparing Theorem 7.4 to Theorem 7.5, the latter being a restatement of the above mentioned result of Dilworth and Crawley. In a somewhat different vein we will show that a cer tain type of lattice homomorphism is useful in studying ir redundant representations and the role played by covers in such representations. Adopting the terminology used in (3) 3 we say that a lattice homomorphism .<|>:L-+-L* is a complete join homomorphism if L and L' are complete and for each Sc=L we have *(U S)= U S € o (s). Our principal results concerning such maps are the following. If L is an upper continuous modular lattice then there is an associated upper continuous modular lattice L* which is the "largest" homomorphic image of L (under a complete join epimorphism) possessing no covers. (See Theorem 5.3) Each element of L has an irredundant repre sentation in teirms of meet irreducibles if and only if each element of L* has such a representation. In particular, if L* consists of one element then each element of L has an irredundant representation in terms of meet irreducible ele^ ments. The class of upper continuous modular lattices for which L* consists of one element is large enough to include both the atomic ones, in which each element has an irredun dant representation in terms of completely meet irreducible elements, and those satisfying the ascending chain condition, in which each element has an irredundant representation as the meet of a finite number of meet irreducible elements. 4 2. TORSION ELEMENTS, TORSION FREE ELEMENTS, AND COVERING CONDITIONS We begin by generalizing two concepts from group theory. Definition. An element t of a lattice L is a torsion element if for each a The element 0 is always a torsion element. If 1 is a torsion element, we will call L a torsion lattice. Definition. An element of L is torsion free if no element of L covers it. Here 1 is always torsion free. Let H be a subgroup of an abelian group G, and let L(G) denote the lattice of subgroups of G. H is a torsion element of L(G) if and only if H is a torsion group. H is torsion free in L(G) if and only if G/H is a torsion free group. In fact the duals of the above definitions have counterparts in the theory of abelian groups. H is dual torsion free in L(G) if and only if H is a divisible group. H is a dual torsion element of L(G) if and only if G/H is a reduced group. In general it is not the case that a lattice must possess maximal torsion elements or minimal torsion free elements. Examples are easily constructed by violating the following condition. Definition. A lattice satisfies the lower covering condition if a>-a n b implies au b>-bl The dual of this condition is termed the upper covering 5 condition. If a and b are elements of a modular lattice then the intervals [an b,a] and [b,au b] are isomorphic. Thus a modular lattice satisfies both covering conditions. Our first theorem is a direct generalization of a clas sical theorem of group theory. Theorem 2.1 In a complete lattice which satisfies the lower covering condition there is an element which is both the largest torsion element and the smallest torsion free element. Before proving this we will develop some lemmas. In the remainder of this section it is assumed that the lattice under discussion is complete and satisfies the lower covering condition. . Lemma 2.1 If x is a torsion element of "L, , and [y, z] has no atoms, then y^. Proof. Assume x n y Lemma 2.2 If x is a torsion element and y is torsion free then x^. Proof: Let z«l. Then x,y, and z satisfy the hypo 6 thesis of Lemma 2.1. Lemma 2.3. If T is a collection of torsion elements then U T is a torsion element. Proof; Suppose y^UT and [y/UT] has no atoms. If tfiT, t;^L/T and Lemma 2.1 implies y^t. Then y^UT. Hence y=UT and UT is a torsion element. Lemma 2.4. If t is a torsion element of L and t'>-t then t' is also a torsion element. Proof; Suppose y^t' and [y,t'] has no atoms. By Lemma 2.1, y^t. But y^t since t't. Hence y=t' and t' is a torsion element. We next investigate the torsion free elements. Lemma 2.5. If F is a collection of torsion free elements then H F is also torsion free. Proof. Suppose c>-nF. Let f be any element of F. If cnf=nF then c>-c n f and the lower covering condition implies fuc>-f, contrary to the hypothesis. Hence cnf=c and c<,f. Thus c^HF - We now prove Theorem 2.1. Let T denote the collection of all torsion elements of L and let t=UT. Lemma 2.3 implies t is the largest torsion element of L. Similarly, if F de notes the collection of all torsion free elements and f=nF then, by the preceding lemma, f is the smallest torsion free element of L. Lemma 2.2 implies t^f. If t Lemma 2.4, t' is a torsion element, contrary to the maximality 7 of t. Hence t=f. Note that the dual of Theorem 2.1 states that in a lattice which satisfies the upper covering condition there is an ele ment which is both the largest dual torsion free (divisible) element and the smallest dual torsion (reduced) element. 8 3. ATOMIC LATTICES AND UPPER CONTINUITY Suppose that L is the lattice of subgroups of an abelian torsion group. If xgL then, since every subgroup of a tor sion group is a torsion group, the sublattice [0,x] is a tor sion lattice and therefore x is a torsion element of L. Hence each element of L is a torsion element. A lattice has this property if, and only if, it is atomic. Any atomic lattice is a torsion lattice but the converse is not true. The first theorem of this section will give conditions which are suffi cient to imply that a torsion lattice is atomic. Before stating these conditions it will be convenient to state a property of upper continuous lattices which we will use fre quently. Lemma 3.1. Suppose that in an upper continuous lattice there are elements a, b, and c such that b A c=a. Then there is an element m^c which is maximal with respect to the property b n m=a. Proof. Let P={p|p^p, bA p=a}. P is not empty since it contains c. Suppose that Sep and S is a chain. Then US^c and, by upper continuity, bn U S=U (brt s)=U a=a. The S C & Sep union of a chain of elements of P is thus also an element of P. P therefore contains maximal elements. Theorem 3.1. An upper continuous torsion lattice which satisfies the upper covering condition is atomic. Proof: Suppose that x lemma implies the existence of an element z which is maximal with respect to the property y n z=x. If z=l we have x=z f» y= iny=y, contrary to x element there is an element z' such that z'>-z. Now z^z u (y n z•)^z '. If z=z u (y n z') then yn z'^z and x=y n z^ n z' ^ n z or yoz'=x, contrary to the maximality of z. Thus z u(y A z')=z' >-z. By the upper covering condition, ynz'>-ynz*n z=y n z=x. That is, y^ n z' >- x. Hence the lattice is atomic. Suppose L is a complete lattice satisfying the lower covering condition. Then if x is any element of L the lattice [x, 1] also satisfies the lower covering condition. Theorem 2.1 then implies the existence of an element t(x) which is both the largest torsion element and the smallest torsion free element of the lattice [x, 1]. Then t(x) must also be the smallest torsion free element of L above x. Since t(x) is a torsion element of [x, 1] the lattice [x, t(x)] is a torsion lattice. Moreover, we have the following as an immediate consequence of the preceding theorem. Lemma 3.2. If x is an element of an upper continuous lattice satisfying both covering conditions then the sub- lattice [x, t(x)]is atomic. In such lattices the torsion free elements have a regu larity property. Lemma 3.3. If x and y are elements of an upper con tinuous lattice satisfying both covering conditions then 10 t(x n y)=t(x) n t(y) . Proof. Since t(xny) is the smallest torsion free element above xn y» Lemma 2.5 implies t(xn y)^t(x) n t(y). We will complete the proof by showing that the opposite in equality also holds. Let P={p[x^^t (x), pny^t(xny)}. P is not empty since x€ P. Suppose that SCP and S is a chain. Then x^US<,t(x). By upper continuity, ((JS)ny =Ug g g (s n y) 4t(xny). Thus the union of a chain of elements of P is again in P. Therefore P contains maximal elements. Let x' be such an element. Suppose x' Similarly, there is am element y' which is maximal with respect to the properties y Theorem 3.2. Let L be an upper continuous lattice 11 which satisfies both covering conditions. If the torsion free elements of L form a chain, L is modular. Proof. A lattice is not modular if, and only if, it contains elements a, b, and c such that a (ii) a'0 c'=b'n c'. (iii) a'uc'=b'vc'. (iv) a'nc' (v) a'nc' Let P be the partially ordered set consisting of all those elements p such that a n c^^a u c, p=(b vp) A (c V p), and (aup)A b=a. P is not empty since anc€P. Suppose that SCP and S is a chain. Let s' =US. Clearly anc^s'^auc. By upper continuity, (bus') n (cus')= (buUg^g gSi) o (c u '-'gj e s®2^ •['-'s^ e s I''" n f«s <-4^ 6 st 'b u =1).0 g s(c U Sj)] U 12 Consider the term (buSj^) n (c u s^). If s^^sj* (b u s^) n (c o s^ ^XbVSglm (cus2)=S2=SiUS2. Similarly, if (bv s^) n (cu Sg)ls^^=s^U Sg. Thus s'^(b u s') n (eus') lUg^ S2ÊS^®1^ ®2^~^s 6 S®~®' s'=(bus')n(cvs'). Also, (a vs') n b=(a n ^ gS) n b=[ Ug ^ g (a u s)] n b= Ug^g [(a u s) n b] =CJ a=a. Therefore the union of a chain of elements of S w O P is again in P and P contains maximal elements. Let m be such an element and let a'=aum, b'=bum, and c'=cvm. Clearly a'^b*. If a'=b', aum=bum and a=(au m)nb =(bV m)n b=b, contrary to a Similarly, if a'nc'=c' we have a'=a'vc'=b'uc' so that a'^b', again contradicting a' Suppose there is an m' in L such that a'^'>-a'r» c'. We will show that m'€ P. Since m'>-a'hc'=m, this will con tradict our choice of m. Clearly anc^'^auc. Also, since m- Since au m'=au m, bum'=bvauin'=buau m=b u m. Consequently, (bu m' ) n (c U m') = (bu m) n (cum'). Now n (cU m)^'. If 13 m' n (cum)=m' we would have m'^cu m and therefore m'4(c u m) n (b u m)=m- (bum) n (cum')4cum. But this implies that m'>-m =(b u m) n (cum)^(bu m) n (c u m') = (b u m') n (cum') , a contradiction. Therefore c u m -< (c u m) u [(bum) n (cum')] = cum'. By the lower covering condition, (bum) n (cum') (cum) m [(bum) n (cum')] = (bum) A (cu m)=m. That is, (bum)n (c u m') = (b um') O (cum')>-m. Since (bum') n (cum') ^m' >- m we have m' =(bu m') n (cum'). Thus m' € P and we have the desired contradiction. Similarly, if we assume the existence of an element m* such that c'^m'>-a'n c', we can show that m'e P, contradicting the maximality of m. Since m - 4 (b u m) n m'4(b u m)n(c u m)=m. That is, m'n (b um)=m- 4(bum)n (cu m)=m, a contradiction. Therefore bum-<(bum) [(bum') n (cumg= bum'. By the upper covering condition, (bum') n (c um) >- (b u m)n[(bum') n (c u m)] = (b u m) n (c u m) «=m. Since c u m'=c u m, (bum') n (cum')>-m. Then (bum') m (cu m') ^'>-m implies m'= (bu m') n (cum'). We next show that 14 (aum')nb=a. Since m=m'n (b c/m)^m'n (aom)^ we have m' n (a u m)=in- (a u m) - to b vm'>-bum. Therefore au m=(au m) u [b n (a u m') ] -< a u m'. Then b n (a v )±a v m and b n (a u m* )^ o (a u m). Since b n (au m'}^ n (au m) we have b n (aum')=b n (au m)=a. Thus m'€ P and we again have the desired contradiction. We have therefore shown that a.', b', and c* satisfy conditions (i) through (v). Since the torsion free elements of L form a chain, either t(a')4t(c') or t(c') Similarly, t(c')4t(a') contradicts condition (v). We there fore must conclude that the lattice does not contain elements a, b, and c such that a which satisfies both covering conditions is atomic and modular. Proof. Theorem 3.1 implies atomicity. Theorem 3.2 implies modularity since a torsion lattice contains only one torsion free element. 15 This corollary is a slight extension of a result due to Dilworth and Crawley (2). 16 4. COMPLETE JOIN HOMOMORPHISMS Before proceeding to the representation theory it is necessary to establish several elementary facts concerning complete join homomorphisms. These homomorphisms will play a prominent role in the theory which is to follow and it will therefore be convenient to adopt certain notations con cerning them. If 9 is an equivalence relation in a lattice L we let denote the natural mapping L+L/6. For x€L let Vg (x)= U{p € Llp=x(i*!Dde) }=? 0'{p €L|(|)g (p) = Ogfx)}. We will use C(L) to denote the collection of congruence relations 8 for which the natural mapping *g:L+L/8 is a complete join epimorphism. The elements of C(L} may be characterized in the following fashion. Lemma 4.1. Suppose L is a complete lattice, ScL, l€S, S is closed under intersections, and letting w(x) = n{s€S I s^x} the equation w(x n y)=w(x) A w(y) holds for all X, y£L. If a relation 8 is defined in L by saying x=y (mod e) if and only if w(x)=w(y) then 8€C(L), w(x)=Vg(x), and S={x € L|VQ (x)=x}. Conversely if 8€C(L) then S={x € L|Vg (x)=x} is such a set and VQ(x)fw(x) for each x€L. Supplement. For 8 €C(L) the following hold. (i) ({,Q(VQ(X))= (iii) Vg(VQ(x))=Vg(x). (iv) If x^ then Vg(x)^^ (y). 17 (V) Vg (xn y)=VG(x) o VQCy). (vi) *Q(x)<*g(y) if and only if Vg (x) «j)0(nQ)=nq£Q<^0(q). Proof. It is clear that 6 is an equivalence relation. For x€L let S^={s€Sl s^}. Then xçw(x) = DS^. Note that if x^ then Ds^^OSy, and w(x)4w(y). Since S is closed under intersections, w(x) € w(w(x))=w(x) . Thus w(x)=x (mode) and VG(x)^w(x). If y=x (mode) then y =w(x). Consequently we also have VG(x) =x}. If x=x* (mod e) and y=y* (mod e) then w(xH y) = w(x)nw(y)=w(x')nw(y')=w(x'n y'3 andxoy-:- ny* (mod e). Thus 0 preserves finite intersection We next verify that it preserves unions. If 0'' -aen gQW(q) and hence w(UQ) Thus (.J, g QW(q)4w(UQ)and w( UG ^QW(q))4W(W(UQ ))=w (U Q) • consequently w(UQ)=w(Ug^ Qw(q)). If x=x' (mod e) and y=y' (mod 6) then w(x U y)=w(w(x) u w(y))=w(w(x* ) U w(y• )) =w(x'uy'). Thus 0 is a congruence relation and the natural map * g :L-*'L/0 is a lattice epimorphism. Note that if x € L then *g(x)= *g(w(x)). Suppose ACL. For a € A we have a<,UA and *Q(a)<#Q(UA). Thus $ g(U A) is an upper bound of the set {<(» 0(a) }a Ç Suppose that b is also and upper bound of this set and let b'£L be such that 4)'g(b')=b. For a € A we have 18 *0 ), *Q(a)m*g(b')=*g(a), *Q(anb')=*Q(a), Vg(anb') =Vq (a), w(anb')=w(a), w(a)nw(b')=w(a), and a^w(a)^w(b' ). Thus UA^w(b') and (U A)4((ig (w(b') )=«j>0 (b')=b. Hence (UA) is the least upper bound of {•e ^ a* That is, (|)g (UA)= ^ ^(|)q (a). Each subset of L/8, being the image of a subset of L under i|ig, has a least upper bound. Since L/6 is bounded below by *g(0) it follows that L/6 is complete. Thus *g:L+L/8 is a complete join epimorphism. Before proving the converse portion of the lemma we will verify the properties listed in the supplement. (i) Suppose xeL and P={p|pix(mod 9)}. Since pre serves unions we have *g(Vg(x))= *g(lJp)= lJpgp*p(p)= Up ep*9 (ii) If «|»Q (x)=(|»Q (y) then x=y (mod 6), {P1P=X (mod 0)} ={p|p-y (mod 6)}, and therefore Vg(x)=VQ(y). Conversely, if Vg(x)=VQ(y) then (frg (x)=(frg (Vg (x))=4>q (v^ (y))=^q (y). (iii) Since ()>q (Vg(x) )=<|)q (x), (ii) implies Vg (Vg (x)) = VG(X). (iv) If x^y then «|)g (y)=(|»g (xu y)=(j)g (x) u Vg (x) U Vg (y)4Vg (y). It follows that Vp(x)çyg(y). (v) By (iv), Vg (xO y)iVQ (x) and Vg(xny) 1 Vg (xny). Thus Vg(xn y)=Vg(x) n Vg(y). (vi) If Vg(x) =*Q(y). Conversely, if 4g(x)<^g(y) then *Q(%)= =*0 (xny), Vg (x)=Vg (xn y)=VQ (x) n Vq (y), and Vq (x)iVg (y). Thus *G(x)<4Q(y) if and only if Vg(X)^VQ(y). Then (ii) implies the desired result. (vii) For g 6 Q we have . flQ^q and hence v^ ( Q 0)17^ (q) =q. Therefore Vg (0 0)400* Since the opposite inequality also holds we have Yg (D0)= DO. If q € Q then q^OO and *g(q)>4g(f)0)' Thus Hq^Q *g(q)>4g(nô). Let p be such that ^'g (P)= g Q *g(q). Since •g (q)^4'g (p)^«frQ (H 0) we have Vg (q)^Vg (p)^Vg (flQ) and therefore q>yg(p)>/10. ThenDo ^Vg(p)^ Or nO=Vg(p), and «|>g(0 0)=«J»g (Vg (p))=«J> q (p) "Oj6Q +e(9'' We now complete the proof of the lemma. Suppose e éC(L) and S= |s|Vg{s)=s}. For x€L let S^= {seSjs^}. Since Vg (x) 6 S^ we have x^Os^^Vg (x). Then Vg (x)4Vg ( 0 S^^) V0(nSx)= ns^. Thus Vg (x)= n{s € Sjs^}. Then (v) and (vii) imply S has the desired properties. The basic feature of complete join epimorphisms which makes them relevant to the study of meet representations is the following. If L is a complete lattice and e €.C(L) then the set {x€ LlVg(x)=x} determines a meet subsemilattice of L which is isomorphic to L/6 when the latter is considered as a meet semilattice. It is therefore possible to obtain meet representations of some elements of L from representa tions in L/e. The following two results illustrate this. 20 Lemma 4.2. If 0€C(L> and v.(x)=x then x is meet irreducible if and only if *g(x) is a meet irreducible element of L/e. Proof. Suppose x is meet irreducible and *g(x) =yj^ny2* Let x^ and Xg be elements of L such that *g(x^) = y^ and *g(x2)=y2. Then (x^^n ^^1^ ^ *6 (x2)=yin yg =*Q(x). Hence Vg (x^^n X2) =Vg (x). Since Vg(x^n X2) = VG (XJ^) n VG (X2) and VQ(X)=X we have x=VG (x^) n VG (XJ). The irreducibility of x implies that either Vq(Xj^)=X or VgfXg) =x. If Vg(x^)=x then <|»g (x)=<|)g (Vg (x^) )=(J>g (x^)=y^. Thus *g(x) is meet irreducible. Suppose *g(x) is meet irreducible and x^n X2=x. Then •e(x)=(|.0(Xin X2) = = Vg(x)=x. Since we also have x^^x it follows that x^=x. Thus X is meet irreducible. Lemma 4.3. Suppose e E C(L) and v (x)=x. Then 8 *g(x) has an irredundant representation in terms of meet irreducible elements of L/e if and only if x has an irre dundant representation in terms of meet irreducibles, x^Or, such that v (r)=r for each r € R. V Proof. Suppose x= (1R is a representation with the stated properties. Since v (r)=r for each rcR we have V 4.(x)=*.(nR)=(l_,g*.(r). The preceding lemma implies u V r € it 8 {(j»e(r)}rçR is a collection of meet irreducible elements of L/e. Suppose € R and ^ ^*g (r)= 0^ ^ *g(r). Then 21 •QCDRj-OrgR Hj. g r-Tq "^0 ( ) • Conse quently Vq { 0 R) =Vg ( n ) and A R= n(R-rQ), contrary to the irredundancy of the representation of x. Thus OrgR 4g(r) for each r^ € R and we have an irredundant representation of $g(x). Conversely, suppose *g(x) has an irredundant represen tation in terms of meet irreducible elements of L/0, say •e (x)=n^^q^. For each o € A let (q^) where q^ is any element of L such that *a(q2)=q_'V u 0( Since v.(r_)=r_OA A for each o e A we have *6(H „ ^ ^ (r^)= n.« A*,(Vg (q;)= ri„ « = « a'S<.=*6 • consequently Ye'n. Then e A'^a='e "^o € A'^a'=^9 ""=*• The preceding lemma implies that each r^ is meet irreducible. Suppose «^«A and *= 0^ ^ Then A-ao^.' = a-ao*0 n, e contrary to the irredundancy of the representation of Another property of complete join epimorphisms which makes them valuable when one is dealing with upper continuous lattices is the following. Lemma 4.4. If L is upper continuous and 6 € C(L) then L/e is also upper continuous. Proof. Suppose x € L/ e, ScL/e, and S is a chain. Let x' be such that $ (x')=x and for each s € S let s' be such V that $g(s')=s. Then {Vg(s')}ggg is a chain in L. The upper continuity of L and the join preserving nature of imply 22 xnUs = *0(x' ) n g g^e(s' ) = •j(x')n U^^g^j(Vg(s')) = •s(x')fl •j[UsesVj(s')] = *e[x'n Uses^e's'l] = •efUses'*'" **(3'))] = Uggg$,(x'n Vg(%')) - Usss[*,(x')n+e(v,(s')l] = Uses [x"*e(s'l] Thus L/0 is upper continuous. Since C(L) is a subset of the lattice of all congruence relations of L, it is a partially ordered set. That is, 0^402 if x=y (mod 0^^) implies x=y (mod Gg). Two results regarding this partial ordering which we will have need of later are given below. Lemma 4.5. If 0^^,02 6C(L), 0^ V. (x)=x. ®1 Proof. Since 0^462' {y|y=x (mod 0^)}c{y|y=x(mod Oj)}. Then x^v. (x) (1 ) ® 2 * 23 (ii) There is a complete join epimorphism ijcLg-*-Lg such that (iii) There is a function such that Proof. (iii) implies (ii). Suppose SCL^. For each ses let s'6L be such that. = *' UL«s+l(='''=*+l''Js«sS')=*2(UsfsS')=(Js<= S+Z's'' = (J^^g*4^(s')= Ug2g*(s). In a similar fashion it can be shown that 4* preserves finite intersections. Thus * is a complete join homomorphism. If x € Lg there exists x'€ L such that *2(x')=x. Then **^(x')=x. Thus i|; is an epimorph ism. (ii) implies (i). If x=y (mod 8^) then *2(x)=**i(x) =**l(y)=*2(y) x=y (mod Gg). Thus (i) implies (iii). If xeL^^ let ij; (x)=({»2 (*') where x' is any element of L such that *^(x')=x. If x" is such that 4i^(x")=x then, since «l'2 ^*"^"*^2 ^ * Thus * is a function from to Lg and by definition **^(x')=*2(x') for each x'€ L. We will also have occasion to discuss suprema and infema of subsets of C(L). Lemma 4.7. C(L) is a complete lattice. Proof. We will first show that any subset of C(L) has a greatest lower bound. Suppose { Galoga/^CCL). Let S be the collection of all those elements which can be ex pressed as the intersection of a subset of {x|Vg (x)=x for (X 24 some oca}. Then l€S and it is clear that S is closed under intersections. Note that each element of S may be expressed in the form Hcx^/ia _ ,x^ where v.o^aa (x: )=x_. For xeL let S x= {s € s Is^x}. Consider the element w(x)=fls^. For each 9^ we have v. (x) € S and therefore w(x)=ns„ we have Vj (x) Then there exists 8 6 C(L) such that S=|xe L |Vq(x)=x} and w(x)=Vg(x). Note that Vg(x)= ^ (x). We will show a that x=y (mod 6) if and only if x=y (mod 6^) for each 6 . It then follows that @ =(1^ _ ,8 . If x€L then x The following result describes the join of a subset of C(L) in the case whep L is upper continuous. Lemma 4.8. Suppose L is an upper continuous lattice. 25 {*a}aeA<^C(L), and Then {X|VQ(X)=X} = {x|Vg (x)=x for each 0^}. a Proof. Let S= {x1vq(x)=x for each 0^}. Then 1€S a and it is clear that S is closed under intersections. For XGL let S^=«{s€S|s^} and let w(x)= NS^. Note that since S is closed under intersections we have w(x)€ S^. This implies that Vg (w(x))=w(x) for each 0^. It also implies a w(w(x))= w(x) for each x€L. In order to verify that S determines an element of C(L) we must show that w(xn y) =w(x)n w(y) for all x,y€L. If a<^ then (Is^ and w(a)4w(b). Since xn y fore have w(xn y)4w(x) n w(y). Let P= {P1X^4W(X), pfiy 4w(xny)}. P is not empty since x€ P. Suppose QCP and Q is a chain. Then upper continuity implies pnUQ = Ug ^ qPH q4 Uq^jQW(xn y)=w(xn y). Since x^UOl w(x), (JQ^P. Thus P contains maximal elements. Let m be such an element. Since x^m^wCx) we have x^^v. (m)4V- (w(x))= a a =w(x) for each 0^. Also, Vq (mny)^VQ (w(x n y))=w(xn y). a a Furthermore w(xo y)^v (m.ny)=v. (m) n Vg (y)^. (m)ny. o a a a Thus V. (m) € P for each 0 . By the maximality of m we ®o * therefore have v^ (m)=m for each 0^. Since x^m^wCx) it 01 follows that w(x)^w(m) w(x)n y4w(xny). similarly, we may take m' maximal with respect to the properties y^m'^wCy) and w(x) n ra'4w(xn y). Again it must be the case that m'=w(y) and therefore w(x) n w(y)4w(xn y). Since the opposite inequality also 26 holds we have w(x) n w(y)=w(xn y). Then Lemma 4.1 implies the existence of 8 € C(L) such that w(y)=Vg(y) and S= {x|VG(x)=x}. For each x€L and each 8^ we have X4VG.(x) <,w(x) and therefore VQ (X)=X (mod 6). If x=y (mod 8^) a then xHVg (x)"Vg (y)=y (mod 6). Thus for each o g A. a a Suppose 0'eC(L) is such that 8 '^8^^ for each a ÇA. Then Lemma 4.5 implies VG (v^, x))=VG,(x) for each a€A. Hence a Vg,(x)eS^ and Vq,(x)^w(x)Consequently x=w(x) (mod 8'). If x=y (mod 8) then x=w(x)=w(y)=y (mod 8'). Thus 0'^8. We therefore have 8= ^ 27 5. TWO HOMOMORPHISMS In some lattices the torsion free elements determine a congruence relation of the type discussed in the preceding section. Theorem 5.1. If L is an upper continuous lattice satisfying both covering conditions then there exists ecC(L) such that Vg{x)=t(x) for each xeL. Proof. In Lemma 4.1 take S to be the collection of all torsion free elements of L. By Lemma 2.5, S is closed under intersections. Then w(x) = (!{s € S| s^}=t (x) . By Lemma 3.3, t(xn y)=t(x) n t(y) for all x,y eL. Thus there exists 0 €C(L) such that Vg (x)=w(x)=t(x) for each x€ L. Henceforth we will use L^ to denote the lattice L/8, where 0 is the congruence relation given by the above theorem. One can easily characterize the upper continuous lattices satisfying both covering conditions for which L*" consists of one element. In such a lattice one has *g(x)=$g(l) for each x e L. Hence t(x)=t(l)=l for each xe L and the only torsion free element is 1. Thus L is a torsion lattice and Corollary 3.1 implies that L is an atomic, modular, upper continuous lattice. It is readily seen that the converse also holds. More generally, L may thought of as the lattice which is obtained by collapsing each of the covers in L upward. 28 We illustrate by constructing in a simple case. Let L be the collection of sequences (x^fXgfXg,...) where each is either 0 or 1. If these are ordered lexicographically then L is à chain and therefore satisfies the hypothesis of the above theorem. The only covers which occur in L have the form (Xj^/...f X^/ IfOfOfOy...) (x^ f •• • r fOfl/lfl^...) « If 0 is the congruence relation of the preceding theorem than a 6 class either contains only one element or consists of a pair of elements having the above form. Then 0 is precisely the equivalence relation one would use to obtain the numbers of the closed real unit interval as limits of binary approximations. Thus is isomorphic to that interval. Incidentally, this same example serves to show that compact generation, unlike upper continuity, need not be preserved by a complete join epimorphism. L is compactly generated, the compact elements being those sequences which are zero from some point on. However the only compact element of is 0. Theorem 5.2. Suppose is a complete join epimorphism, where and Lg are upper continuous lattices satisfying both covering conditions. Then there is a t t complete join epimorphism such that the diagram 29 1» -» L- is commutative. Proof. We first show that for each x€L. Let P={p|x^p^t(x) r Suppose Sep and S is a chain. It is clear that x^US<.t(x). The map ^2^ is a complete join epimorphism since both and ill are. Then 42*(US)=Ug^g*2*(s)= lJgg = (JI (x)=*_*(x) and US€P. Thus P contains maximal S € O Z 6 elements. Let m be such an element. If m Lemma 3.2, there exists m' such that m- *(m')=*(m) we have contrary to the maximality of m. Thus *(m)<*(m'). Suppose *(m) If m=m u (m* n y') we have m^'n y' and *(m)>^(m')n*(y') =i|)(m* )n y=y, contrary to *(m) =m' and *(m' )=* (m) U [*(m')n*(y')]=*(m)U(*(m')n y) =*(m)uy=y. Thus *(m')>-*(m). Hence <|/(m)- Then But by the nature of *2 we have 4»2'l'(m) =«^2 ('I'(m) )]. Therefore =*2*(x)' contrary to the maximality of m. We are thus led to conclude that m=t(x) and ®1' @2 6C(L) be the congruence relations corresponding to the 30 homomorphisms (|>^ and respectively. If x=y (mod 6^) then t(x)=t(y), and x=y (mod Bg). Thus Lemma 4.6 then guarantees the existence of the desired homomorphism. Corollary 5.1. Suppose is a complete join epimorphism, where is atomic, modular, and upper continu ous. Then Lg is also atomic, modular, and upper continuous. Proof. Lg is modular since any homomorphic image of a modular lattice is again modular. Lemma 4.4 implies that Lg is upper continuous. Then by the preceding theorem t t there is a complete join epimorphism ^/iL^+Lg. Since L^^ is atomic, L^ consists of one element. Since t|» ' is an epimor- phism it follows that Lg consists of one element. Thus Lg is atomic. Suppose now that L is an upper continuous modular lat tice and e ec(L). Then L/e is again an upper continuous modular lattice. Hence the torsion free elements of L/0 determine an element of C(L/0). The mapping L-»-L/e->(L/0)^ is a composite of two complete join epimorphisms and is there fore also a mapping of this type. Thus there is an element of C(L) corresponding to the mapping L+(L/0)^. We will use 0^ to denote the element of C(L) obtained from 0 in this man ner. Thus L^=L/0^. It is clear that 0^>4 for each 0 € C(L). Lemma 5.1. Suppose L is an upper continuous modular lattice, 0^,02 €C(L), and 0]^<02. Then 31 Proof. Consider the following diagram, L^- >L/e£ L/02 XL/eg)^ Since 81I82' Lemma 4.6 implies the existence of the complete join epimorphism ip making the left half commutative. Then Theorem 5.2 guarantees the existence of the complete join epimorphism *' such that the right half commutes. Again applying Lemma 4.6, the existence of the mapping i|;' im plies Lemma 5.2. Suppose L is an upper continuous modular lattice, 8 € C(L), and VQ(X)=X. Then IF and only if Ogfx) is a torsion free element of L/8. Proof. Let *' denote the mapping L/e+(L/8)^. Suppose Ogfx) is torsion free. If x=y (mod 8^) then *'*g(x) =*'*@(y) and therefore t(*g(x))=t(*g(y)). Hence Ogfy) <.t(*0 (y) )=t((|>Q (x))=(J>g (x). Then VQ(y)4VQ(x) and we have (y)4VQ (x)=x. Thus Vgt(x)= U{y|y=x (mod 8^) }=x. Suppose <(>g(x) is not torsion free. Then 4g(x) Let ycL be such that (y)=t («|>g (x)) . Since «|»q (y) >«|»g (x) we have Vg (y)^Vg (x)=x. But Vg(y)^ x since q (Vg (y)) =«|»q (y) =t(*g(x))>*g(x). Note that Vg(y)=x (mod 8^) since •••Q (Vg(y))=^'(j.Q(y)=(|.'(t(*e(x))=4.'(|>g(x). Thus v^t(x) = U{P1P=X (mod 8^)}>VG (y)>x. Lemma 5.3. If L is an upper continuous modular 32 lattice and 8€C(L) then the following are equivalent. (i) 0^=0. (ii) Each element of L/0 is torsion free. (iii) L/G contains no covers. (iv) Between any two comparable elements of {x|Vg(x)=x} there is a third. Proof. (ii) implies (i). Let x be any element of L. Since v^(Vq(x))=Vg(x) and *g(Vg(x)) is torsion free the preceding lemma implies Vg^.(Vq (x))=Vg (x). Since x=Vg (x) (mod 0) and 0<0^ we have x=Vq(x) (mod 0^), and hence Vgt(x) =VQt(V0(x)). Thus Vgt(x)=Vg(x). If x^sxg (mod 0^) then Vg(Xi)=V0t(Xi)=Vgt(X2^~^0^*2^ and X1HX2 (mod 0). Therefore 0=0^ (i) implies (ii). Let x be any element of L/0 and let x'€L be such that *g(x')=x. Since Vg|.(Vg(x)) = Vg(Vg(x))=Vg(x) the preceding lemma implies (jig(Vg(x')) is torsion free. But Lemma 5.4. Suppose L is an upper continuous modular lattice and 0 cC(L). Then there is a unique 0* €C(L) such that the following hold. (i) 0*>0 33 (ii) (8*)t=8* (iii) If 8^>8 and 9^=6 then Proof. Let S={0' €C(L)l0'^6,(6')^=e'}. S is not empty since it contains the unit congruence relation. Let 0*=ns. Then 8*>6. For each 8' €S we have 8* =0*. If 0^ is as in (iii) then 0^ € S and therefore ®1^^S=0*.. 0* is unique since (iii) forces any two ele ments satisfying (i) and (ii) to be equal. The congruence relation o* will be of particular in- , terest to us and we will adopt special notation pertaining to it. Let L* denote the lattice L/0*, let tp* denote the homomorphism *^^:L+L/0*=L*, and let v*(x)=v^^(x) for xeL. The lattice L* may be characterized in the following fashion. Theorem 5.3. If L is an upper continuous modular lattice then: (i) L* contains no covers. (ii) If *:L+L' is a complete join epimorphism and L' contains no covers then there is a complete join epimorphism i|»':L*-^L' such *=*'$*. Proof. Since (0*)^=0*, Lemma 5.3 implies L* con tains no covers. Let 0 denote the congruence relation in L determined by the mapping *:L+L'. Then 0 6C(L) and L/0=L*. Since L* contains no covers. Lemma 5.3 implies 0=0^. It follows from Lemma 5.4 that 8>0*. Then Lemma 4;6 implies 34 the existence of the desired mapping ij;' ;L/0*->L/e. Theorem ,5.4. If *:L^+L2 is a complete join epi- morphism, where and Lg are upper continuous modular lattices, then there is a complete join epimorphism such that the diagram is commutative. Proof. By (i) of the preceding theorem con tains no covers. Then the complete join epimorphism satisfies the hypothesis of (ii) of that theorem. Thus there is a complete join epimorphism such that We conclude this section by showing that two important types of lattices have the property that L* consists of one element. If L is a modular lattice which satisfies the ascending chain condition then L* consists of one element. Assume the contrary. Then v*(0) then L* consists of one element. Since L is atomic t consists of one element and 0 is the unit congruence rela tion, Since 0^4(0*)^=0* it follows that 0*=0^ and L* con sists of one element. 36 6. MEET IRREDUCIBLE ELEMENTS In this section we list some elementary properties of meet irreducible elements in order to gain some insight in to the manner in which such elements can arise. Lemma 6.1. An element x is completely meet irre ducible if and only if there exists x' such that x-cx'^ for each y>x. Proof. Suppose x is completely meet irreducible. Let S={y|y>x} and let x*= S. By hypothesis x >x. If y>x then y € S and y^ns=x'. Therefore x' >- x. Conversely, if x' is an element with the stated properties then fl{y|y>x} = n{y|y^' }=x' >-x. Thus X is completely meet irreducible. There is a simple criterion for determining which meet irreducible elements are completely meet irreducible. Lemma 6.2. A meet irreducible element is completely meet irreducible if and only if it is not torsion free. Proof. By Lemma 6.1, no completely meet irreducible element is torsion free. Conversely, suppose x is meet irreducible and x'>-x. If y>x then x^x'Dy^x'. By hypo thesis x'n y^x. Hence x'n y=x' and y^x'. Thus Lemma 6.1 implies x is completely meet irreducible. In an upper continuous lattice the following may be used to obtain meet irreducible elements above a given element. Lemma 6.3. Suppose x' >x, x is meet irreducible in 37 terms of elements of [x,x'], and r is maximal with respect to the property r A x'=x. Then r is meet irreducible. If x is completely meet irreducible in terms of elements of [x,x'] then r is completely meet irreducible. Proof. Suppose y^>r, yj^ny2=r. Then x=r n x'= (y^n ygin x'= (y^n x') n (y^n x'). Here y^n x', ygO x' €[x,x'] and, by the maximality of r, y^n x'>x and y2nx*>x. But this contradicts the hypothesis regarding X. Hence r is meet irreducible. The second assertion may be verified in a similar fashion. As an immediate consequence of this we have; Lemma 6.4. If x'>-x and r is maximal with respect to the property r x'=x then r is completely meet irreducible. For modular lattices there is a converse to Lemma 6.3. Lemma 6.5. Suppose that r is a meet irreducible element of a modular lattice, x pletely meet irreducible then x is completely meet irreduc ible in terms of elements of [x,x']. Proof. . Suppose s>r and x'n s=x. By modularity, (rux')0 s=r u(x'n s)=rux=r. Since r is meet irreducible it follows that r Vx'=r. Then x'=x'n r=x, a contradiction. Thus r is maximal with respect to the stated property. Now suppose that ' » and y^^H y2=x* Let 38 Pl=yi n (Yg u r) and 9^=72 n (y^U r). Then p^n p^ =yi A (YgA r) n Yg n (y^^n r)=y^^n Y2=^- Also, since 4x* we have (p^^U P2) D r=x. By modularity, r U p^ =r u[y^ n (y^u r)] = (ru y^^) n (r U y^). Similarly, r u p^ =(r u y^^) n (rU yg). Thus r u P2=r U P2=r up^ u Pg. Conse quently (p^U r) n (Pj^ OP2)= (P3^UP2 <-/r) n (p^U P2)=PiU P2» On the other hand, modularity implies (p^ur) D (Pg^ u P2) =Pj^.U [rn(Pj^ u P2)] =Pi U x=p^. Hence p^^p^ u P2 and P2=Pi n Pg =x. Then r=rux=rup2=r u[y2 n (r u y^)] = (r uy2) n (r Uy^^). It follows from the irreducibility of r that either r U y^ = r or rOy2=r. If, say, rUyj^=r then yj^=rnyj^=x. Thus X is meet irreducible in terms of elements of [x,x']. If, in addition, r is completely meet irreducible then, by Lemma 6.1, there exists r' such that r'>-r. Then r 4r u (x* n r' ) . If r u (x* O r * ) =r then x*nr'=r n(x'n r' ) =x'n r=x, contrary to the maximality of r. Hence rU(x'nr') =r' r. The upper covering condition implies x'Or' >-r n (x'o r' ) =x' n r=x. Since x - irreducible in terms of elements of [x,x'], by applying Lemma 6.2 in the sublattice [x,x'] we see that r is com pletely meet irreducible in terms of elements of this interval. The following result will be needed later. Lemma 6.6. If x is a torsion free element of a modular lattice and x=r)R is an irredundant representation in terms of meet irreducible elements then each element of 39 R is torsion free. Proof. Suppose r €R is not torsion free and r'>-r. By Lemma 6.2, r is completely meet irreducible. Then since n(R - r)>x and r n n(R - r)=x the preceding lemma implies that X is completely meet irreducible in terms of elements of [x,n(R - r)]. Applying Lemma 6.1 in the sublattice [x, D(R - r)] we see there must exist x* such that x- <.n(R - r) r contrary to the hypothesis that x is torsion free. Hence each element of R is torsion free. Theorem 6.1. Suppose that L is a modular lattice in which each element has an irredundant representation in terms of meet irreducible elements. Then for each x Since x'>x and x*nr=x. Lemma 6.5 implies that x is meet irreducible in terms of elements of fx,x']. Now suppose L is upper continuous and x 3.1 there exists an element z which is maximal with respect to the property z n y=x. If z=l then x=lny=y, contrary to x Since r A (z'n y)=zny=x. Lemma 6.5 implies that x is meet irreducible in terms of elements of [x,z'ny]. Thus x'=z'ny is the desired element. The above result concerning meet irreducible elements is analogous to the following one, basically due to Dilworth and Crawley, dealing with completely meet irreducible ele ments (2). Theorem 6.2. Suppose that L is a modular lattice in which each element has an irredundant representation in terms of completely meet irreducible elements. Then L is a torsion lattice. If, in addition, L is upper continuous then L is atomic. Proof. Suppose x is torsion free and x=nR is an irredundant representation in terms of completely meet irreducible elements. By Lemma 6.6 each element of R is torsion free. But Lemma 6.2 implies that each element of R is not torsion free. Hence R is empty and x= (1 R=1. Thus L is a torsion lattice. Theorem 3.1 then yields the second assertion. 41 7. IRREDUNDANT REPRESENTATIONS In this section we will obtain various conditions for the existence of irredundant representations in upper continuous lattices. The technique used is an extension of one developed by Crawley (1). In attempting to obtain a representation of an element x of a lattice L we will con sider quadruples of the form (x,y,R,8) where; (i) y€L, RCL, and 0€C(L). (ii) Each r€ R is meet irreducible. (iii) x^yoOR. (iv) y is maximal with respect to (iii). (v) x We may define a relation among these quadruples by saying (x,yj,R^,8i) (vii) yi4y2'^i^ ^2'®14®2* (viii) (yify2'*2"^l'G2) admissible quadruple. (ix) V (r)=r for each r €R~-R,. 0^ ^ i. Thus if (x,y^,R^,ej^)4(x,y2/R2'®2^ then the representa tion x=y2nr)R2 a refinement of the representation xsy^nnR^. We first verify that this relation is a partial 42 ordering. It is clear that it is reflexive and anti symmetric. We will show that it is transitive. Assume (Xfyi'Rl'Glli/XfygfRgfGg) {Xry2'R2'®2^=^*'^3'^3'®3^ • Then (y]^fy2'^2"^l'®2^ and (y2fy2/R3-R2'®3^ are admissible, Vg (r)-r for each reRg'^i' and Vg (r)-r for each reR^-Rg. We must verify that (yi'yg'Bg^Ri'Gg) is admissible and that Vg (r)=r for each reR^-R^. The latter assertion is clear in view of Lemma 4.5. Condition (iii) is satisfied since y^nOCRj-R^^)=y3nn(R3-R2)n fl(R2-Ri)=y2nn(R2-Ri)=yi. Suppose y'^3 and y'nfl(R3-Ri)=yi. Since [y'nfl(R3-R2)] nn(R2"^l)-y'^ 0(^3-^2)=yi and y'nn(R3-R2^^3 the maximality of y^ implies y ' a n(I^3~I^2^~^2* ^hen the maximality of y3 implies y'zy^. Thus y3 is maximal with respect to the property y3n n(R3"Ri^~yi (yi'y3'*3~*l'G3) satisfies (iv). If r € R3-R2 then y2'^y3 and by the maximality of y2'y3nfl(R3-R2-r)= [y3nfl(R3-R2-r)] I' Suppose r eR2"'^i yjnn(R3-R3^-r)=yj^. Then y^=y2nO(R3-R2)n n (R2-R3^-r)=y2nn(R2"^l"^^' contrary to the admissibility of (yi'y2'*2"*l'^2)' Thus yi {(x,x,0,O)}. It is clear that the union of a chain of elements of P is again an element of P. Thus P contains maximal elements. Suppose that M-{(x,yQ,RQ,8^)}^^_^ is a maximal chain. We will show that '^CL e € A".' g aO.) e M. It then follows that this is a maximal quadruple. Let R ^R and 8= U ^ ,8 . Since M is a chain the =V/ ^€ A a a 6 A a definition of the partial ordering implies that is also a chain. If (x,y^,R^,8^) GM and r € R-R^ then there exists (x,yg,Rg,8g)€M such that (x,y^,R^,6^) " n 6 A n « A n R,)nn(R-R.)] = U nn(R-Rg)]= g Lemma 3.1 implies the existence of an element y^ ^ which is maximal with respect to the property yn nR=x. We first show that 44 (x,y,R,8) is an admissible quadruple. By our choice of y,R, and 0, conditions (i) through (iv) are satisfied. If (x,y•'a ,R a ,ra )€M then •*'=^ay^ and r^• a for each r € R-Ra . Thus ynn(R-R^)^^. Also y n D (R-R^^) n nR^=y n nR=x. Since (x,y^,R^,6^)^(x,x,0,O) the maximality of y^ implies y nn(R-Rg)=yg^ for each a 6 A. If r 6 R there exists (x,y^,R^,8^)€M such that rGR^. Then ynn(R-r)= , y nn(R-Rg)n n (R^-r)=y^nn(R^-r)>x. Thus (x,y,R,8) satisfies (v). If (x,y^,R^,0^)€ M and r£ R-R^ then there exists (x,yg,Rg,8g)€M such that (x,y^,R^,8Q) V. (r)=r. That is, if r6 R-R then v. (r)=r. Since 0a (X u a Vq (y^)=y^ and y^=ynn(R-R^) we have y^=VQ (y„)= a a Vq (ynn(R-Rjjj) )=Vq (y)nvQ (fl(R-R^))=Vg (ylnflfR-R*). a a a a Since Vg (y)^ and Vg (y) nnR=VQ (yyn/l(R-Rg)nflRa=yan/lRa a a a =x our choice of y implies v. (y)=y. Since v. (y)=y for a a each 0^, Lemma 4.8 implies Vg(y)=y. Thus (x,y,R,8) satis fies (vi). We next show that (x,y,R,8)>,(x,y^,R^,6^) for afiA. Since we have already seen that v.(r)=r for each a it suffices to ®a show that (y ,y, R-R ,8) is an admissible quadruple. Con- a a ditions (iii) and (vi) have already been verified. We next establish (iv). Suppose y'^ and y'nfl(R-R^)=y^. Then y'onR=y'0 n(R-H )ndR =y oHr =x and by our choice of y a a (X a we have y'=y. It only remains to verify (v). Suppose re R-R and ynOCR-R -r)=y . Then ynOCR-r) a a o 45 =ynOR^ n n (R-Rg^-r) =y^ n =x, contrary to the admissibility of (x,y,R,8). Thus y n n(R~Rjj-r) >y^ for each rgR-R^ and (x,y,R,6)^(x,y^,R^,0^). It follows from the maximality of M that (x,y,R,0) CM. It is then clear that this is a maximal quadruple such that The proof of the lemma will be complete when it has been shown that f^fy) has the stated properties. Suppose anb=*g(y) where a is meet irreducible and b>*g(y). By Lemma 4.4, L/0 is upper continuous. Lemma 3.1 then im plies the existence of a b^^ which is maximal with respect to the property aAb^=*Q(y). Let a' and b* be such that *g(a')=a and *g(b')=b2. Let y'=Vg(b') and r'=Vg(a'). By Lemma 4.2, r' is meet irreducible. It is easily verified that (y,y',{r'},8) is an admissible quadruple. Since Vg(r')»=r' it follows that (x,y,R,e)4(x,y* ,Rv{r'},0) . Since (y')=bj^>(j)g(y) ,Vq (y')=y', and VQ(y)=y, we have y'>y. Hence (x,y,R,6)<(x,y',Rv{r'},0), contrary to the maximality of (x,y,R,0). Thus there cannot exist elements a,b with the stated properties. Suppose O^fy) is not tor sion free. Then there exists b such that b>-<|»g(y). Let a be maximal with respect to aAb=*g(y). Lemma 6.4 implies that a is completely meet irreducible. But this contra dicts the property of *g(y) which was just verified. Thus *g(y) is torsion free. Theorem 7.1. An upper continuous lattice L has the property that each element possesses an irredundant repre 46 sentation in terms of meet irreducible elements if and only if for each y Proof. Suppose y€L, y Suppose X6L and take 8^=0 in the preceding lemma. If (x,y,R,8) is a maximal quadruple then 8=0, x=ynnR, x ties is 1 we conclude that y=l. Then x= (iR is an irredun dant representation. One can use the above to obtain conditions which are sufficient for the existence of irredundant representa tions. Theorem 7.2. Suppose that L is an upper continuous lattice and for each x meet irreducible in terms of elements of [x,x']. Then each element of L has an irredundant representation in terms of meet irreducibles. Proof. For x . theorem. By Lemma 3.1 there exists an element r which is maximal with respect to the property r/)x*=x. Lemma 6.3 implies r is meet irreducible. The preceding theorem there 47 fore yields the desired conclusion. The analogous result dealing with completely meet irreducible elements is the following. Theorem 7.3. If L is an upper continuous torsion lattice then each element of L has an irredundant repre sentation in terms of completely meet irreducible elements. Proof. If x' >—X then x is meet irreducible in terms of elements of [x,x']. The preceding theorem therefore implies that each element of L has an irredundant represen tation in terms of meet irreducible elements. Suppose y= OR is such a representation. Irredundancy implies l^R. Hence no element of R is torsion free. Lemma 6.2 then im plies that each element of R is completely meet irreducible. Combining Theorem 7.2 with Theorem 6.1 one obtains the following. Theorem 7.4. If L is an upper continuous modular lattice then the following are equivalent. (i) Each element of L has an irredundant representation in terms of meet irreducible elements. (ii) For each x meet irreducible in terms of elements of [x,x']\ (iii) Whenever x is meet irreducible in terms of elements of [x,x']. Similarly, by combining Theorem 7.3 with Theorem 6.2 one has: Theorem 7.5. If L is an upper continuous modular 48 lattice then the following are equivalent. (i) Each element of L has an irredundant representa tion in terms of completely meet irreducible elements. (ii) For each x (iii)Whenever x We now state our principal result. Theorem 7.6. If L is an upper continuous modular lattice then each element of L has an irredundant represen tation in terms of meet irreducible elements if and only if each element of L* has such a representation. Proof. Suppose each element of L* has such a representation and let x be any element of L. In Lemma 7.1 take 0Q=O* to obtain a maximal quadruple (x,y,R,8), 8^0*. Since Vg(y)=y and Thus (x,y,R,6^) is admissible.and the maximality of (x,y,R,G) implies 8=8^. It follows that 8=0* and v*(y)=y. By hypo thesis 4*(y) has an irredundant representation in terms of meet irreducible elements of L*. Then Lemma 4.3 im plies that y has an irredundant representation in terms of meet irreducibles, y= D S^ where v*(x)=x for each s€ S. Then (y,l,S,0*) is admissible and (x,l,RyS,0*)^(x,y,R,o*)• Hence x= is a representation of the desired type. Now suppose each element of L has such a representation and let x be any element of L*. Let x'£ L be such that 49 **(x')=x and let y=v*(x'). Suppose y= DR is an irredundant representation in terms of meet irreducibles. We will show that v*(r)=r for each reR. Since 4*(y)=4*(x')=x it then follows from Lemma 4,3 that x= ^ _**(r) is a representa- r c K tion having the desired properties. Let S={e' eC(L) VG,(r)=r for each reR} and let e=Us. Lemma 4.8 implies ees. By Lemma 4.3, (y)= €R*8 irredundant representation of (y) in terms of meet irreducible ele ments of L/e. Since e^O* we have 046^40*. Since v*.(y)=y it follows from Lemma 4.5 that Vg(y)=Vq^(y)=v*(y)=y. Then Lemma 5.2 implies ^^(y) is a torsion free element of L/6. By Lemma 6.6, {(|)g (r) ^ is a collection of torsion free elements of L/8. Since v-(r)=r for each reR it follows V from Lemma 5.2 that Vgt(r)=r for each reR. Then 6^6 S and 6=/ls>y^. Hence 8=8^=0* and v*(r)=r for each reR. More generally, one can show that if L is an upper continuous modular lattice, 8€C(L), and 848*40*, then each element of L/0 has an irredundant representation in terms of meet irreducible elements if and only if each element of L/0' has such a representation. As an immediate consequence of Theorem 7.6 we have the following. Theorem 7.7. If L is an upper continuous modular lattice and L* consists of one element then each element of L has an irredundant representation in terms of meet irreducible elements. 50 Note that this implies the existence of representa tions in modular lattices which satisfy the ascending chain condition and in atomic, upper continuous, modular lattices. We conclude this section with an example of a very well behaved lattice which contains elements having no irredundant representation in terms of meet irreducible elements. Let L be the collection of all monotone non- decreasing functions of the closed real unit interval into itself. If suprema and infema are defined pointwise then L becomes a complete lattice. One can then easily verify the following remarks. (i) L inherits very strong regularity properties from the unit interval, including modularity and upper continuity. (ii) The meet irreducible elements of L are those functions which have the form; f(x)=yQ if X possess an irredundant representation in terms of meet irreducible elements. 51 8. APPLICATIONS We will now give some results which sometimes allow one to infer the existence of irredundant representations in an upper continuous modular lattice from properties of a set of generators of the lattice. Such results have natural applications in compactly generated modular lattices. In particular, these results apply to module theory and we will give examples in this field. Theorem 8.1. Suppose L is an upper continuous modular lattice, G CL, UG=1, and each of the lattices Lg= [0,g],g€G, has the property that each element possesses an irredundant representation in terms of meet irreducible elements of L . Then each element of L ha? an irredun- 9 dant representation in terms of meet irreducible elements. Proof. We will show that L satisfies (ii) of Theorem 7.4. Suppose x Lg has an irredundant representation in terms of meet irre ducible elements of Lg, Theorem 7.4 (ii) implies the exis tence of y sucn that xng [xog,y] and [x,xuy] are isomorphic. Hence x The converse to the above is somewhat trivial. For 52 suppose L is an upper continuous modular lattice and g€L. If each element of L has an irredundant representation in terms of meet irreducible elements then, by Theorem 7.4 (iii), whenever x Then the lattice L^=[0,g] satisfies condition (ii) of Theorem 7.4. Therefore each element of L has an irre- 9 dundant representation in terms of meet irreducible ele ments of Lg. As an immediate consequence of Theorem 8.1 we have the following. Corollary 8.1. Suppose L is a compactly generated modular lattice and each of the lattices L^=[0,k], k com pact, has the property that each element possesses an irre dundant representation in terms of meet irreducible ele ments of L^. Then each element of L has an irredundant representation in terms of meet irreducible elements. Corollary 8.2. Suppose L is a compactly generated modular lattice in which every element which lies below a compact element is compact. Then each element of L has an irredundant representation in terms of meet irreducible elements. Proof. If k is a compact element such that each « element below k is also compact then the ascending chain condition holds in [0,k]. Thus Corollary 8.1 yields the 53 desired conclusion. An example of a lattice satisfying the hypothesis of the preceding corollary would be the lattice of submodules of a left module over a left noetherian ring. Results analogous to those given above also allow one to conclude that a lattice L has the property that L* con sists of one element. Theorem 8.2. Suppose L is an upper continuous modular lattice, Gcl, UG=1, and each of the lattices l = 9 [0,g], gcG, has the property that L* consists of one element. Then L* consists of one element. The proof requires the following lemma. Lemma 8.1. An upper continuous modular lattice L has the property that L* consists of one element if and only if the only subset S of L which has the properties listed below is the set {l}. (i) 16 S. (ii) S is closed under intersections. (iii) Letting w(x)= n{s € S| s^x} the equation w(xny) =w(x)nw(y) holds for all x,y 6 L. (iv) Between any two comparable elements of S there is a third. Proof. Lemma 4.1 and Lemma 5.3 imply that {v*(x)}^^^ is a set having these properties. If {l} is the only such set then v*(x)=l for each XCL. Consequently **(x)=**(v*(x))=**(!) for each x€L and L* consists of one 54 element. Conversely, suppose L* consists of one element and 5 has the listed properties. By Lemma 4.1 there exists 6 eC(L) such that S={Vq (x) ^ Lemma 5.3 implies that L/0 contains no covers. It follows from Theorem 5.3 that there is a complete join epimorphism *:L*+L/8. Since L* consists of one element it must therefore be the case that L/0 contains only one element. Then for each xCL we have *g(x)=*Q(l) and therefore Vg (x) =Vg (1) =1. -phus S={1}. Proof of Theorem 8.2. For g e G let S={gnv*(x) It can be verified that S is a subset of L_ which satisfies 9 the conditions listed in the preceding lemma. Since L* consists of one element it follows that S={g}. Hence g^v*(x) for each g€ G and each x€L. Then V*(X)^UG=1 and v*(x)=l for each x€L. Consequently **(x)=**(v*(x)) = Then L* consists of one element. Corollary 8.2 may be sharpened somewhat. Corollary 8.4. Suppose L is a compactly generated modular lattice in which every element which lies below a compact element is compact. Then L* consists of one element. Proof. Since each of the lattices L^=[0,k], k 55 compact, satisfies the ascending chain condition the pre ceding corollary yields the desired conclusion. We will conclude by giving two examples of applications in module theory. By a ring R we will mean a commutative ring with identity. Suppose M is an R module, is a collection of submodules of M, and A = 0. Then M a G A a has a canonical subdirect product representation, M c _,M/P . If P is a meet irreducible submodule then TT'a 6 A ' o o the rank of M/P^ is one. If the meet representation of 0 is irredundant then each of the factors M/P^ in the sub- direct product representation has a non-zero intersection with M. Thus results regarding the existence of representa tions in the lattice of submodules of M imply the existence of various types of subdirect product representations of M. The lattice of submodules of a module M is a compactly generated modular lattice. Theorem 8.3. The following are equivalent. (i) Each ideal of the ring R has an irredundant representation in terms of meet irreducible ideals. (ii) If M is any R module then every submodule of M has an irredundant representation in terms of meet irreducible submodules. Proof. Let L(R) denote the lattice of ideals of R and if M is an R module let L(M) denote the lattice of submodules of M. The first condition is a trivial conse- 56 quence of the second since R is itself an R module and the submodules of R are its ideals. Suppose R satisfies (i) and M is any R module. For mg M let (m) denote the submodule of M generated by m. Then I={r € R|rm=0} is an ideal of R and there is a module isomorphism between (m) and R/I. Hence the intervals [0,(m)]CL(M) and fI,R]c:L(R) are iso morphic. Since each element of [I,R] has an irredundant representation in terms of meet irreducible elements it follows that each element of the lattice L(^^=[0,(m)] has such a representation. The desired conclusion then follows from Theorem 8.1 by taking G={(m) ^ Theorem 8.4 The following are equivalent. (i) The lattice L(R) of ideals of the ring R has the property that L(R)* consists of one element. (ii) If M is any R module then the lattice L(M) of submodules has the property that L(M)* consists of one element. Proof. Again it is the case that (i) is a trivial consequence of (ii). Suppose (i) holds and M is an R module. If me M there is an ideal I of R such that the interval [I,R] of L(R) is isomorphic to the interval [0,(m)] of L(M). The only subset of L(R) satisfying the conditions listed in Lemma 8.1 is the set {R}. In particu lar, the only such subset of [I,R] is {R}. Hence the only subset of the lattice satisfying those condi 57 tions is {(m)}. Therefore L*consists of one element for each m g M and the desired conclusion is obtained from Theorem 8.2 by taking G={(m)}^^^. Theorem 8.2 implies that a ring which is a direct sum of noetherian rings satisfies condition (i) of the above theorem. 58 9. BIBLIOGRAPHY Crawley, P. Decomposition theory for nonsemimodular lattices. Amer. Math. Soc. Trans. 99: 246-254. 1961. Dilworth, R. P. and P. Crawley. Decomposition theory for lattices without chain conditions. Amer. Math. Soc. Trans. 96: 1-22. 1960. Szasz, G. Introduction to lattice theory. 3rd rev. ed Academic Press, New York, N.Y. 1963. 59 10. ACKNOWLEDGMENTS The author wishes to express his gratitude to his men tors, Dr. T. J. Head and Dr. B. Vinograde, and to his wife Evelyn for their invaluable advice, assistance, and inspira tion.